All categories

2 years ago

What is the area of this parrelogram

Mathematics

1 answer:

Dmitrij [34]2 years ago

7 0

Answer:

no parallelogram, mr clean thus cannot answer.

You might be interested in

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

9 months ago

The value of y varies directly with x, and y=10 when x=4. Find x when y=14.

damaskus [11]

$y=10\ \ \ and\ \ \ x=4\\\\
y=14\ \ \ and\ \ \ x=?\\\\
Making\ ratio:\\
10-4\\14-x\\\\
Cross\ multiplication:\\
10x=4*14\\
10x=56\ \ \ |Divide\ by\ 10\\x=5,6\\\\solution\ is\ x=5,6.$

5 0

2 years ago

EXAMPLE 5 If F(x, y, z) = 4y2i + (8xy + 4e4z)j + 16ye4zk, find a function f such that ∇f = F. SOLUTION If there is such a functi

Valentin [98]

If there is such a scalar function f, then

$\dfrac{\partial f}{\partial x}=4y^2$

$\dfrac{\partial f}{\partial y}=8xy+4e^{4z}$

$\dfrac{\partial f}{\partial z}=16ye^{4z}$

Integrate both sides of the first equation with respect to x :

$f(x,y,z)=4xy^2+g(y,z)$

Differentiate both sides with respect to y :

$\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}$

$\implies\dfrac{\partial g}{\partial y}=4e^{4z}$

Integrate both sides with respect to y :

$g(y,z)=4ye^{4z}+h(z)$

Plug this into the equation above with f , then differentiate both sides with respect to z :

$f(x,y,z)=4xy^2+4ye^{4z}+h(z)$

$\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}$

$\implies\dfrac{\mathrm dh}{\mathrm dz}=0$

Integrate both sides with respect to z :

$h(z)=C$

So we end up with

$\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}$

7 0

2 years ago

Which of the following states the conditions that the graph meets?

Sladkaya [172]

B is correct
I just took it

8 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

slega [8]

Let r(x,y) = (x, y, 9 - x^2 - y^2)

So, dr/dx x dr/dy = (2x, 2y, 1)

So, integral(S) F * dS
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx
= 28/5 + 40/9 - 1/4
= 1763/180

6 0

2 years ago